Course 5: Rank-Nullity & Isomorphisms

1. Rank-Nullity Theorem

Theorem 5.1: Rank-Nullity Theorem

Let $T: V \to W$ be a linear map between finite-dimensional vector spaces. Then:

\dim(V) = \dim(\ker T) + \dim(\text{im } T)

Equivalently: $\dim(V) = \text{nullity}(T) + \text{rank}(T)$ , where $\text{nullity}(T) = \dim(\ker T)$ and $\text{rank}(T) = \dim(\text{im } T)$ .

Proof of Theorem 5.1:

Let $\dim(\ker T) = k$ and let $\{u_1, \ldots, u_k\}$ be a basis of $\ker T$ .

Extend this to a basis $\{u_1, \ldots, u_k, v_1, \ldots, v_m\}$ of $V$ .

We claim $\{T(v_1), \ldots, T(v_m)\}$ is a basis of $\text{im } T$ .

Spanning: Any $w \in \text{im } T$ equals $T(v)$ for some:

v = \sum_{i=1}^k \alpha_i u_i + \sum_{j=1}^m \beta_j v_j

Then:

w = T(v) = \sum_{i=1}^k \alpha_i T(u_i) + \sum_{j=1}^m \beta_j T(v_j) = \sum_{j=1}^m \beta_j T(v_j)

Independence: If $\sum \beta_j T(v_j) = 0$ , then:

T\left(\sum \beta_j v_j\right) = 0

So $\sum \beta_j v_j \in \ker T$ , hence $\sum \beta_j v_j = \sum \alpha_i u_i$ .

By independence of the full basis, all coefficients are zero.

Therefore $\dim(\text{im } T) = m$ and:

\dim(V) = k + m = \dim(\ker T) + \dim(\text{im } T)

∎

Corollary 5.1: Consequences

For a linear map $T: V \to W$ with $\dim V = n$ :

$T$ is injective iff $\dim(\ker T) = 0$ iff $\dim(\text{im } T) = n$
$T$ is surjective iff $\dim(\text{im } T) = \dim W$
If $\dim V = \dim W$ , then $T$ is injective iff surjective iff bijective

Example 5.1: Applying Rank-Nullity

Let $T: \mathbb{R}^5 \to \mathbb{R}^3$ with $\dim(\ker T) = 2$ .

By Rank-Nullity: $5 = 2 + \dim(\text{im } T)$ , so $\dim(\text{im } T) = 3$ .

Since $\dim(\text{im } T) = \dim(\mathbb{R}^3)$ , $T$ is surjective.

Corollary 1.1: Dimension Bounds

For $T: V \to W$ with $\dim V = n$ and $\dim W = m$ :

$\dim(\ker T) \geq n - m$ (at least this many vectors must map to zero)
$\dim(\text{im } T) \leq \min(n, m)$ (image cannot exceed domain or codomain dimension)

Theorem 1.1: Rank-Nullity for Compositions

For $T: U \to V$ and $S: V \to W$ :

\dim(\ker(S \circ T)) = \dim(\ker(T)) + \dim(\ker(S) \cap \text{im}(T))

2. Isomorphisms

Definition 5.2: Isomorphism

A linear map $T: V \to W$ is an isomorphism if it is bijective (both injective and surjective).

If such an isomorphism exists, we say $V$ and $W$ are isomorphic, written $V \cong W$ .

Theorem 5.2: Inverse of Isomorphism is Linear

If $T: V \to W$ is an isomorphism, then $T^{-1}: W \to V$ is also linear.

Proof:

For $w_1, w_2 \in W$ and $\alpha, \beta \in F$ :

T^{-1}(\alpha w_1 + \beta w_2) = T^{-1}(\alpha T(T^{-1}(w_1)) + \beta T(T^{-1}(w_2)))

= T^{-1}(T(\alpha T^{-1}(w_1) + \beta T^{-1}(w_2))) = \alpha T^{-1}(w_1) + \beta T^{-1}(w_2)

∎

Theorem 5.3: Classification of Finite-Dimensional Vector Spaces

Two finite-dimensional vector spaces over the same field $F$ are isomorphic if and only if they have the same dimension.

Proof:

(⇒) If $V \cong W$ via isomorphism $T$ , then $\ker T = \{0\}$ and $\text{im } T = W$ .

By Rank-Nullity: $\dim V = 0 + \dim W$ , so $\dim V = \dim W$ .

(⇐) If $\dim V = \dim W = n$ , choose bases $\{v_1, \ldots, v_n\}$ of $V$ and $\{w_1, \ldots, w_n\}$ of $W$ .

Define $T(v_i) = w_i$ and extend linearly. This gives an isomorphism.

∎

Corollary 5.2: V ≅ Fⁿ

Every $n$ -dimensional vector space $V$ over $F$ is isomorphic to $F^n$ .

Example 5.2: Isomorphism Examples

$\mathbb{R}^2 \cong \mathbb{C}$ as real vector spaces (both 2-dimensional)
$P_2(\mathbb{R}) \cong \mathbb{R}^3$ via $a + bx + cx^2 \mapsto (a, b, c)$
$M_{2 \times 3}(\mathbb{R}) \cong \mathbb{R}^6$ via flattening the matrix

Example 5.3: More Isomorphism Examples

$\mathbb{R}^n \cong M_{n \times 1}(\mathbb{R})$ (column vectors)
$M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$ via $A \mapsto (a_{11}, a_{12}, \ldots, a_{nn})$
$\text{Sym}_n(\mathbb{R}) \cong \mathbb{R}^{n(n+1)/2}$ (symmetric matrices)

Theorem 2.1: Isomorphism Preserves Structure

If $T: V \to W$ is an isomorphism, then:

$\{v_1, \ldots, v_k\}$ is independent in $V$ iff $\{T(v_1), \ldots, T(v_k)\}$ is independent in $W$
$\{v_1, \ldots, v_k\}$ spans $V$ iff $\{T(v_1), \ldots, T(v_k)\}$ spans $W$
$\{v_1, \ldots, v_n\}$ is a basis of $V$ iff $\{T(v_1), \ldots, T(v_n)\}$ is a basis of $W$

Definition 5.3: Automorphism

An automorphism is an isomorphism from a vector space to itself. The set of all automorphisms of $V$ , denoted $\text{Aut}(V)$ or $GL(V)$ , forms a group under composition.

3. Dual Spaces

Definition 5.4: Linear Functional and Dual Space

A linear functional on a vector space $V$ over $F$ is a linear map $\phi: V \to F$ .

The dual space of $V$ , denoted $V^*$ , is the vector space of all linear functionals on $V$ :

V^* = \mathcal{L}(V, F)

Theorem 5.4: Dimension of Dual Space

If $V$ is finite-dimensional, then $\dim(V^*) = \dim(V)$ .

Proof:

$V^* = \mathcal{L}(V, F)$ , so $\dim(V^*) = \dim(V) \cdot \dim(F) = \dim(V) \cdot 1 = \dim(V)$ .

∎

Definition 5.5: Dual Basis

Let $\mathcal{B} = \{v_1, \ldots, v_n\}$ be a basis of $V$ . The dual basis $\mathcal{B}^* = \{v_1^*, \ldots, v_n^*\}$ of $V^*$ is defined by:

v_i^*(v_j) = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}

Example 5.3: Dual Basis Example

For the standard basis $\{e_1, e_2\}$ of $\mathbb{R}^2$ :

$e_1^*(x, y) = x$ (extracts first coordinate)
$e_2^*(x, y) = y$ (extracts second coordinate)

Definition 5.6: Dual Map

For a linear map $T: V \to W$ , the dual map (or transpose) $T^*: W^* \to V^*$ is defined by:

T^*(\phi) = \phi \circ T

for all $\phi \in W^*$ .

Theorem 3.1: Dual Basis is a Basis

If $\mathcal{B} = \{v_1, \ldots, v_n\}$ is a basis of $V$ , then $\mathcal{B}^* = \{v_1^*, \ldots, v_n^*\}$ is a basis of $V^*$ .

Proof:

For any $\phi \in V^*$ , write $v = \sum \alpha_i v_i$ . Then:

\phi(v) = \phi\left(\sum \alpha_i v_i\right) = \sum \alpha_i \phi(v_i) = \sum \phi(v_i) v_i^*(v)

So $\phi = \sum \phi(v_i) v_i^*$ , showing $\mathcal{B}^*$ spans. Independence follows from evaluating on the $v_j$ .

∎

Remark 5.1: Double Dual

The double dual $V^{**}$ is the dual of $V^*$ . For finite-dimensional $V$ , there is a natural isomorphism $V \cong V^{**}$ given by $v \mapsto \text{ev}_v$ where $\text{ev}_v(\phi) = \phi(v)$ .

4. Applications of Rank-Nullity

The Rank-Nullity Theorem has numerous applications in solving linear systems, understanding matrix properties, and characterizing linear maps.

Theorem 4.1: System of Linear Equations

For a system $Ax = b$ with $A \in M_{m \times n}(F)$ :

If $\text{rank}(A) = m$ (full row rank), the system is consistent for all $b$
If $\text{rank}(A) = n$ (full column rank), the system has at most one solution
If $\text{rank}(A) = n = m$ , the system has a unique solution for all $b$

Proof:

The system $Ax = b$ is consistent iff $b \in \text{im}(T_A)$ where $T_A(x) = Ax$ .

If $\text{rank}(A) = m$ , then $\dim(\text{im}(T_A)) = m = \dim(F^m)$ , so $\text{im}(T_A) = F^m$ (surjective).

If $\text{rank}(A) = n$ , then $\dim(\ker(T_A)) = 0$ , so $T_A$ is injective (at most one solution).

∎

Example 4.1: Determining Solvability

For $A \in M_{5 \times 3}(\mathbb{R})$ with $\text{rank}(A) = 2$ :

By Rank-Nullity: $\dim(\ker(A)) = 3 - 2 = 1$ .

The system $Ax = b$ has either no solution (if $b \notin \text{im}(A)$ ) or infinitely many solutions (if consistent, since kernel has dimension 1).

Theorem 4.2: Dimension of Solution Space

For a consistent system $Ax = b$ , the solution set is an affine space of dimension $\dim(\ker(A))$ .

Example 4.2: Application to Matrix Factorization

If $A \in M_{n \times n}(F)$ has $\text{rank}(A) = r$ , then $A$ can be factored as $A = BC$ where $B \in M_{n \times r}(F)$ and $C \in M_{r \times n}(F)$ .

This follows from $\dim(\text{im}(A)) = r$ , so we can choose $r$ linearly independent columns for $B$ .

5. Dual Maps and Transpose

Every linear map induces a dual map (transpose) that reverses direction. This fundamental construction connects linear maps with their matrix representations and reveals deep symmetries.

Theorem 5.1: Properties of Dual Map

For linear maps $T: V \to W$ and $S: W \to X$ :

$(S \circ T)^* = T^* \circ S^*$ (composition reverses)
$(I_V)^* = I_{V^*}$ (identity dual is identity)
If $T$ is an isomorphism, then $(T^{-1})^* = (T^*)^{-1}$

Proof:

(1) For $\phi \in X^*$ :

(S \circ T)^*(\phi) = \phi \circ (S \circ T) = (\phi \circ S) \circ T = T^*(S^*(\phi)) = (T^* \circ S^*)(\phi)

(3) Since $T^* \circ (T^{-1})^* = (T^{-1} \circ T)^* = I_V^* = I_{V^*}$ , we have $(T^*)^{-1} = (T^{-1})^*$ .

∎

Theorem 5.2: Kernel and Image of Dual Map

For $T: V \to W$ :

$\ker(T^*) = (\text{im}(T))^0$ (annihilator of image)
$\text{im}(T^*) = (\ker(T))^0$ (annihilator of kernel)
$\text{rank}(T^*) = \text{rank}(T)$

Proof:

(1) $\phi \in \ker(T^*)$ iff $T^*(\phi) = 0$ iff $\phi \circ T = 0$ iff $\phi(w) = 0$ for all $w \in \text{im}(T)$ iff $\phi \in (\text{im}(T))^0$ .

(3) By (1) and dimension formula: $\dim(\text{im}(T^*)) = \dim(W^*) - \dim((\text{im}(T))^0) = \dim(W) - (\dim(W) - \dim(\text{im}(T))) = \dim(\text{im}(T))$ .

∎

Example 5.1: Dual Map in Coordinates

If $T: \mathbb{R}^n \to \mathbb{R}^m$ is given by $T(x) = Ax$ for matrix $A$ , then $T^*$ corresponds to $A^T$ (transpose).

This is why the dual map is also called the "transpose" of a linear map.

6. Annihilators and Double Dual

Annihilators provide a way to understand subspaces through their duals. The double dual gives a natural way to identify a vector space with its "dual of the dual", revealing a fundamental symmetry.

Definition 6.1: Annihilator

For a subset $S \subseteq V$ , the annihilator of $S$ is:

S^0 = \{\phi \in V^* : \phi(s) = 0 \text{ for all } s \in S\}

Theorem 6.1: Annihilator is a Subspace

For any $S \subseteq V$ , $S^0$ is a subspace of $V^*$ .

Proof:

If $\phi, \psi \in S^0$ and $\alpha \in F$ , then for any $s \in S$ :

(\alpha \phi + \psi)(s) = \alpha \phi(s) + \psi(s) = 0

So $\alpha \phi + \psi \in S^0$ .

∎

Theorem 6.2: Dimension of Annihilator

If $W$ is a subspace of a finite-dimensional $V$ , then:

\dim(W^0) = \dim(V) - \dim(W)

Proof:

Let $\{w_1, \ldots, w_k\}$ be a basis for $W$ , extended to a basis $\{w_1, \ldots, w_k, v_1, \ldots, v_n\}$ for $V$ .

Then $\{v_1^*, \ldots, v_n^*\}$ (from the dual basis) is a basis for $W^0$ , so $\dim(W^0) = n = \dim(V) - k$ .

∎

Definition 6.2: Double Dual and Evaluation Map

The double dual $V^{**}$ is $(V^*)^*$ . For each $v \in V$ , define the evaluation map:

\text{ev}_v: V^* \to F, \quad \text{ev}_v(\phi) = \phi(v)

This gives a map $\text{ev}: V \to V^{**}$ defined by $\text{ev}(v) = \text{ev}_v$ .

Theorem 6.3: Natural Isomorphism V ≅ V**

For finite-dimensional $V$ , the evaluation map $\text{ev}: V \to V^{**}$ is an isomorphism.

Proof:

$\text{ev}$ is linear: $\text{ev}(\alpha v + u)(\phi) = \phi(\alpha v + u) = \alpha \phi(v) + \phi(u) = (\alpha \text{ev}(v) + \text{ev}(u))(\phi)$ .

Since $\dim(V) = \dim(V^*) = \dim(V^{**})$ , it suffices to show injectivity.

If $\text{ev}(v) = 0$ , then $\phi(v) = 0$ for all $\phi \in V^*$ . If $v \neq 0$ , extend to a basis and use the dual basis to get a contradiction. So $v = 0$ .

∎

Example 6.1: Annihilator Example

In $\mathbb{R}^3$ , let $W = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (xy-plane).

Then $W^0 = \{\phi : \phi(x, y, 0) = 0\}$ . If $\phi(x, y, z) = ax + by + cz$ , then $\phi \in W^0$ iff $a = b = 0$ .

So $W^0 = \{\phi : \phi(x, y, z) = cz\}$ is 1-dimensional, and $\dim(W) + \dim(W^0) = 2 + 1 = 3 = \dim(\mathbb{R}^3)$ .

Theorem 6.4: Double Annihilator

For a subspace $W \subseteq V$ , $(W^0)^0 = W$ (under the natural identification $V \cong V^{**}$ ).

Frequently Asked Questions

Why is the Rank-Nullity theorem so important?

It's the fundamental constraint on linear maps. It tells us that 'dimension lost' (kernel) plus 'dimension gained' (image) equals the starting dimension. Everything about existence and uniqueness of solutions follows from this.

What does 'isomorphic' really mean?

Isomorphic spaces are 'the same' for all linear algebra purposes. They have identical algebraic structure—same dimension, same types of subspaces, same linear maps. Only the 'names' of elements differ.

Why is V ≅ Fⁿ so important?

It means every finite-dimensional space can be studied using coordinates. Abstract theorems about V translate to concrete matrix computations in Fⁿ. This is the bridge between abstract and computational linear algebra.

What's a linear functional?

A linear map from V to the base field F. It assigns a scalar to each vector, linearly. Examples: evaluation at a point, integration, trace of a matrix.

How does Rank-Nullity connect to matrix rank?

For a matrix A representing T, rank(A) = rank(T) = dim(im T) = column rank = row rank. The nullity equals the number of free variables, which is n - rank(A).

Work These Ideas by Hand — Then Check Yourself

RREF Calculator — row-reduce any matrix to reduced row echelon form step by step
Matrix Multiplication Calculator — multiply matrices and check dimension compatibility

Rank-Nullity & Isomorphisms Practice

10

Questions

0

Correct

0%

Accuracy

1

If

T: \mathbb{R}^5 \to \mathbb{R}^3

and

\dim(\ker T) = 2

, what is

\dim(\text{im } T)

?

Easy

Not attempted

2

If

T: \mathbb{R}^4 \to \mathbb{R}^4

is surjective, is it injective?

Medium

Not attempted

3

Can a linear map

T: \mathbb{R}^3 \to \mathbb{R}^5

be surjective?

Medium

Not attempted

4

If

T: V \to W

is an isomorphism, what is

\dim(\ker T)

?

Easy

Not attempted

5

Are

\mathbb{R}^2

and

\mathbb{C}

isomorphic as real vector spaces?

Medium

Not attempted

6

If

\dim V = \dim W

, is every linear

T: V \to W

an isomorphism?

Easy

Not attempted

7

What is

\dim(V^*)

if

\dim(V) = n

?

Easy

Not attempted

8

The dual of a linear map

T: V \to W

is

T^*: ? \to ?

Medium

Not attempted

9

If

\{e_1, e_2\}

is a basis of

V

, what is

e_1^*(e_2)

?

Easy

Not attempted

10

If

\dim V = 5

and

\dim(\ker T) = 2

, what is

\dim(\text{im } T)

?

Medium

Not attempted